0580 · 7.1Transformations

Transformation basics

0:00A triangle can slide across a grid, turn around a point, flip across a line, or change size. These movements may produce similar-looking pictures, but they follow different rules. Understanding those rules lets you predict exactly where every vertex will go. By the end, you will be able to recognise, describe and draw all four transformations, including combinations. A transformation maps an original shape onto an image. The original shape is often called the object, and the image is the result.

0:28Corresponding vertices are matching corners before and after the transformation. We will label original vertices P, Q and R, and their images P prime, Q prime and R prime. You need to read coordinates confidently before starting. The x coordinate gives horizontal position, while the y coordinate gives vertical position. The origin is where both coordinates equal zero. When drawing an image, transform each vertex separately, then join the matching vertices in the original order.

0:58Reflection, rotation and translation preserve lengths and angles, so the image has the same size and shape as the object. Enlargement preserves angles and scales every length by the same factor. Its image therefore has the same shape, although its size may change. Always use a ruler for every straight edge.

Translations using vectors

1:15A translation slides a shape without turning or flipping it. Every point moves the same horizontal distance and the same vertical distance. We describe that movement using a column vector. Its top component gives the horizontal movement, and its bottom component gives the vertical movement. The column vector with x component negative three and y component two means three units left and two units up. Negative horizontal movement goes left, while positive vertical movement goes up.

1:43These components describe changes in position, not the coordinates of the destination. When a vector has a letter name, the exam prints that letter in bold. Let us work through the translation shown. For P, subtracting three from its x coordinate gives negative one. Adding two to its y coordinate gives three. Its image is therefore the point with x equal to negative one and y equal to three. For Q, 4 − 3 gives an x coordinate of one.

2:11Its y coordinate becomes 1 + 2, which equals three. For R, the horizontal calculation gives 2 − 3, or negative one. Vertically, 3 + 2 = 5, so R prime has x equal to negative one and y equal to five. Plot those three points and join them with a ruler. Every side keeps its original length and direction because all vertices received the same movement. To recognise a translation from a diagram, compare corresponding vertices.

2:39Their horizontal changes must agree, and their vertical changes must also agree. To find its vector, subtract an original coordinate from the matching image coordinate. For P, −1 − 2 gives negative three horizontally. The vertical change is 3 − 1, giving two. A complete description is a translation by the column vector with x component negative three and y component two.

Reflections in mirror lines

3:02A reflection flips a shape across a straight mirror line. Each image point lies on the opposite side, the same perpendicular distance from that line. Perpendicular means meeting at a right angle. The mirror line therefore cuts the segment joining a point to its image exactly in half. Imagine folding the paper along the mirror line. Each original vertex would land on its matching image vertex. That explains why distances must be measured straight across the mirror line, not along a slanting route.

3:29Any point already on the mirror line stays in its original position. Reflect our original triangle in the line x = 1. P is one unit to the right because 2 − 1 = 1. Its image lies one unit left, so 1 − 1 gives x equal to zero. The y coordinate stays at one because movement across this vertical mirror line is horizontal. Q lies three units right of the mirror because 4 − 1 = 3.

3:57Moving three units left from the mirror gives 1 − 3, or negative two. R has the same x coordinate as P, so its reflected x coordinate is also zero. Its height remains three, giving R prime at x equal to zero and y equal to three. A complete description is a reflection in the line x = 1. Do not describe it as a reflection in the y axis, because that axis is the line x = 0.

4:20To identify an unknown mirror line, find the midpoint between corresponding vertices. The mirror must pass through those midpoints at right angles to the joining segments. A mirror line can also be diagonal. Reflection in the line y = x exchanges the horizontal and vertical coordinates. The point with x equal to two and y equal to five becomes the point with x equal to five and y equal to two.

4:43The diagram shows equal perpendicular distances on opposite sides, which explains the exchange. Reflection in the line y = −x also exchanges the coordinates, and it changes both signs. Under it, the point two comma five becomes the point negative five comma negative two. Their midpoint has x equal to negative one point five and y equal to one point five, which lies on that mirror line. A horizontal mirror line, such as y = 2, changes only the y coordinates.

Rotations about a centre

5:12A rotation turns a shape about a fixed centre. Every vertex turns through the same angle in the same direction. Its distance from the centre stays unchanged, as though attached by a rigid arm. For this syllabus, rotations are through multiples of ninety degrees. A complete rotation description needs the angle, direction and centre. Ninety degrees clockwise means turning in the direction followed by clock hands. Anticlockwise is the opposite direction.

5:39A rotation of one hundred and eighty degrees gives the same result in either direction, but the centre still matters. Rotate our triangle ninety degrees clockwise about the point with x equal to one and y equal to one. P starts one unit right of the centre. After the turn, it lies one unit below it. Adding that movement to the centre gives x equal to one and y equal to zero.

6:01Q begins three units right of the centre, because 4 − 1 = 3. A clockwise turn places it three units below. Its image has x equal to one and y equal to 1 − 3, giving negative two. We have preserved its distance from the centre rather than moving it around the origin. For R, subtracting the centre coordinates gives one unit right and two units up. Turning clockwise sends the upward part rightwards and the rightward part downwards.

6:27Its new position is two units right and one unit down from the centre. Adding these changes gives x equal to three and y equal to zero. This explains coordinate rotation rules instead of making them mysterious. A clockwise turn converts a relative movement right into down, and up into right. Always measure those movements from the stated centre. Assuming every rotation uses the origin gives the wrong image whenever the centre is elsewhere.

6:52For an anticlockwise quarter turn, a movement right becomes up, and up becomes left. A half turn reverses both parts, so P would land at the point zero comma one. Turning two hundred and seventy degrees clockwise gives the same image as ninety degrees anticlockwise. To find an unknown centre, draw the perpendicular bisectors of the segments joining two pairs of corresponding vertices. The centre is where those bisectors cross.

Enlargements and scale factors

7:16An enlargement works differently because it scales distances from a centre. Draw a straight line from the centre through each original vertex. The scale factor tells you how the directed distance along that line changes. The centre itself stays fixed because its distance from itself is zero. With scale factor two, each image point is twice as far from the centre on the same side. With scale factor 1/2, each lies halfway between the centre and its original point.

7:43An enlargement can therefore make a shape smaller despite its name. Scaling from the wrong centre puts the correctly sized shape in the wrong position. Here, P begins two units right of the centre. Multiplying two by two places its enlarged image four units right, at x equal to five. Using scale factor 1/2 instead gives a distance of one unit, placing it at x equal to two. In both cases, its y coordinate remains one.

8:10A negative scale factor places the image on the opposite side of the centre. Its size is controlled by the size of the factor without the negative sign. Scale factor −1/2 therefore halves every length and reverses each direction from the centre. It does not create negative lengths or reflect the shape in a mirror line. The image looks turned upside down, as if given a half turn about the centre.

8:33Enlarge the triangle shown by scale factor −1/2, centre the point with x equal to one and y equal to one. P lies two units right of the centre. Multiplying that movement by −1/2 gives one unit left. Its image has x equal to 1 − 1, or zero, while y remains one. Q starts four units right, so its scaled movement is two units left. Its new x coordinate is 1 − 2, giving negative one, and its y coordinate stays one.

9:02R begins two units right and four units up. Scaling gives one unit left and two units down, placing R prime at x equal to zero and y equal to negative one. The full description is an enlargement, scale factor −1/2, centre the point one comma one. To recognise an enlargement, compare corresponding lengths to find the scale factor’s size. Extend lines through matching vertices to locate the centre where they meet.

9:29If matching vertices lie on opposite sides of that centre, the scale factor is negative. A full description must include both the centre and the signed scale factor.

Combined transformations and exam checks

9:37Combinations apply one transformation to the image produced by the previous one. Reflect our original triangle in the x axis, then translate it three units upwards. Reflection changes the y coordinates from one, one and three to negative one, negative one and negative three. The x coordinates remain two, four and two. Now add three to each reflected y coordinate. For P, −1 + 3 gives two. Q also finishes at height two, while R reaches −3 + 3, giving zero.

10:08The final vertices are P at two comma two, Q at four comma two, and R at two comma zero. Order matters because the second transformation acts on a different starting image. Translating upwards first would give heights four, four and six. Reflecting afterwards would change those to negative four, negative four and negative six. Label intermediate images carefully so that you do not accidentally transform the original shape twice.

10:34Exams often ask for the single transformation that maps the original object onto the final image. Here each x coordinate stays the same, while each final height equals three minus the original height. The original and final heights always average one point five. The single transformation is therefore a reflection in the line y = 1.5. Never describe a single transformation as two separate steps. Exam questions may ask you to recognise, describe or draw a transformation.

11:02For a description, give the precise information that makes the movement unique. The rotation we worked was ninety degrees clockwise about the point with x equal to one and y equal to one. Saying only that the triangle turns would leave the angle, direction and centre unspecified. Watch for incomplete descriptions, reversed vector components, and distances measured from the wrong centre. A reflection needs its mirror line, not just a statement that the shape flips.

11:28Check that all corresponding vertices follow the same rule. One matching vertex alone does not prove that your transformation maps the whole shape correctly. These grid methods must work without a calculator on Paper two and remain useful on calculator Paper four. Our coordinates are exact, so keep fractional answers exact rather than rounding them. Reflection means equal perpendicular distances across a line, while rotation means equal distances around a centre.

11:55Translation gives every point the same movement, and enlargement scales every directed distance from its centre.